Question

a. use the intermediate value theorem to show that the following equation has a solution on the given interval
b. use the graphing utility to find all the solutions to the equation on the given interval
c. illustrate your answers with an appropriate graph

$x^{3}-5x^{2}+2x=-2,(-1,5)$

evaluate the function f(x) at the right endpoint
the value of the function at the right endpoint is 10
(type an integer or decimal rounded to three decimal places as needed )

why can the intermediate value theorem be used to show that the equation has a solution on (-1,5)?
a. it can be used because f(x)=$x^{3}-5x^{2}+2x$ is defined on (-1,5) and -2 < f(-1)<f(5)
b. it can be used because f(x)=$x^{3}-5x^{2}+2x$ is continuous on -1,5 and the function is defined at x = -1 and x = 5
c. it can be used because f(x)=$x^{3}-5x^{2}+2x$ is defined on (-1,5) and f(-1)<f(5)< -2
d. it can be used because f(x)=$x^{3}-5x^{2}+2x$ is continuous on -1,5 and -2 is between f(-1) and f(5)

b. there is/are a solution(s) to the equation in (-1,5) at x≈
(type integer or decimals rounded to three decimal places as needed. use a comma to separate answers as needed )

Explanation:

Step1: Recall Intermediate Value Theorem

The Intermediate - Value Theorem states that if a function \(y = f(x)\) is continuous on a closed interval \([a,b]\), and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in the interval \((a,b)\) such that \(f(c)=k\). For the function \(f(x)=x^{3}-5x^{2}+2x\), we first find \(f(-1)\) and \(f(5)\).
\[f(-1)=(-1)^{3}-5(-1)^{2}+2(-1)=-1 - 5-2=-8\]
\[f(5)=5^{3}-5\times5^{2}+2\times5=125 - 125+10 = 10\]
The equation is \(x^{3}-5x^{2}+2x=-2\), so we want to find when \(f(x)=-2\). Since \(f(x)=x^{3}-5x^{2}+2x\) is a polynomial, it is continuous on the closed - interval \([-1,5]\), and \(-8=f(-1)<-2<f(5) = 10\), the Intermediate Value Theorem can be used.

Step2: Solve the equation using a graphing utility

We can rewrite the equation as \(x^{3}-5x^{2}+2x + 2=0\). Using a graphing utility (such as a graphing calculator or software like Desmos), we find the roots of the function \(y=x^{3}-5x^{2}+2x + 2\) in the interval \((-1,5)\).
The roots of the equation \(x^{3}-5x^{2}+2x + 2 = 0\) in the interval \((-1,5)\) are \(x\approx - 0.532,1.465,4.067\)

Answer:

a. D
b. \(-0.532,1.465,4.067\)